Can I work backwards from the definition [tex]\intf(x,y)dxdy[/tex] = [tex]\intf(rcos\vartheta,rsin\vartheta)rdrd\vartheta[/tex] and then I can say rdrd[tex]\vartheta[/tex]=dA and so all I have is [tex]\int\intsinydxdy[/tex]? And I'm terrible at bounds so where do I start there?

Staff: Mentor

What exactly is the problem? Is it to evaluate this integral?
[tex]\int_{\theta = 0}^{\pi /2} \int_{r = 0}^{2a cos \theta} cos(2 \theta) r dr d\theta[/tex]
If that's the problem, then all you need to do is evaluate the integral.

If you're trying to convert this to an iterated integral in rectangular (Cartesian) coordinates, which is not at all obvious from the information you've given, then yes dA = dx dy = r dr d[itex]\theta[/itex], and x = r cos [itex]\theta[/itex], y = r sin [itex]\theta[/itex].

The region over which you're integrating is a half-circle of radius 2a, with center at (a, 0) in Cartesian coordinates. Keep in mind, though, that your integral is not the area of this region.

sin instead of cos but yes. my instructions are to convert to Cartesian and then evaluate. But I'm not sure what the integral will be and with what bounds once I convert it...evaluating should be easy.

Staff: Mentor

Re: Converting double integral to Cartesian coordinates

The region over which you're integrating is a half-circle of radius 2a, with center at (a, 0) in Cartesian coordinates.

This half-circle determines the limits of integration, whether as a polar iterated integral or a Cartesian iterated integral. Can you find its equation in terms of Cartesian coordinates?
Also, can you convert sin(2[itex]\theta[/itex]) to Cartesian form?

Can I work backwards from the definition [tex]\intf(x,y)dxdy[/tex] = [tex]\intf(rcos\vartheta,rsin\vartheta)rdrd\vartheta[/tex] and then I can say rdrd[tex]\vartheta[/tex]=dA?

While this is right, I would really really suggest you don't just treat this as a definition but try to understand where it has come from. For a start this will mean it's not something you need to memorise but can just work it out if you ever forget, but more importantly it will help you to understand geometrically what is happening in integrals like this.

What I'm basically asking is, do you understand why it should be [itex]r\;{\rm d}r\;{\rm d}\theta[/itex] and not just [itex]{\rm d}r\;{\rm d}\theta[/itex]?

Another poster has already told you what the bounds mean geometrically, but again I would strongly encourage you to try and sketch it out and make sure you can see this for yourself (of course you should always verify it mathematically after making an initial sketch).

Once you have a sketch, you need to find a way to parametrise the area of integration in Cartesian coordinates. To understand this it may be worthwhile looking through some simpler examples first. A good way to start to get a feel for the way you parametrise areas in integral bounds is to do some exercises in just changing the order of integration with Cartesian double integrals. Once you are happy with this then move on to simple Cartesian<->Polar transformations and get a good feel for that too. I've attached a few examples. Why don't you work through them and then see if this has taught you enough to solve the bounds problem here. If not we can provide additional hints...